I resolve this question by using kinds of iteration approach of subset sum (https://en.wikipedia.org/wiki/Subset_sum_problem), show my code below in Python 2.7. Wondering if any pure mathematical solution?
Source code in Python 2.7,
'''
# Print all integers that can be obtained as
# the sum of cubes of 2 distinct pairs of integers.
# Example: 12 pow 3 + 1 pow 3 = 1729 = 10 pow 3 + 9 pow 3
# x * x * x + y * y * y = num
# k * k * k + z * z * z = num
# (x,y) != (k, z) != (y, x)
'''
def two_sum(numbers, target):
r = 0
i = 0
j = len(numbers) - 1
while i<j:
if numbers[i] + numbers[j] == target:
r += 1
i += 1
j -= 1
elif numbers[i] + numbers[j] > target:
j -= 1
else:
i += 1
return r
def find_number():
cube_numbers = []
result = []
for i in range(100):
cube_numbers.append(i*i*i)
for i in range(1, 10001):
r = two_sum(cube_numbers, i)
if r == 2:
result.append(i)
return result
if __name__ == "__main__":
print find_number()